If you’ve been following the Almagest posts, you’ll recall that we’ve done some work to derive the chord lengths of various angles. But Ptolemy’s goal is to derive the chord length for every angle between 0-180º in $\frac{1}{2}$º intervals. To do that, we’re going to have to develop some new tools using Ptolemy’s theorem on the angles we already know in order to add, subtract, and divide them. These new tools are referred to as corollaries since they come from applications of Ptolemy’s theorem.
The first one comes from the following diagram:
Here $\overline{AD}$ is a diameter. Although Ptolemy didn’t explicitly state it, $\angle{ABD}$ and $\angle{AGD}$ are both right angles. It’s a general rule that if a point lies on the circumference of a circle and connects two other points on the circumference that also lie on a diameter, then it must be a right angle. If you’re having a hard time visualizing that, I actually demonstrated it here, for a specific case, but you can quickly understand how if you move that point that’s the right angle, one increases and the other decreases in a balanced manner always forcing that angle to be 90º. Bringing this up now because we’ll be seeing it a lot coming up.
Moving on, Ptolemy states that if we know $\overline{AB}$ then we know $\overline{BD}$ and similarly if we know $\overline{AG}$ then we also know $\overline{DG}$. Since we just stated that $\angle{ABD}$ and $\angle{AGD}$ are right angles, this comes from the Pythagorean theorem since we have the given side and $\overline{AD}$ is a diameter which Ptolemy gives a length of 120 parts. So that’s two sides of the triangle meaning we can find the third.
But what Ptolemy wants to get at is $\overline{BG}$. Essentially, this is saying, “If we have chord $\overline{AG}$ which is subtends $\angle{ADG}$, we can subtract off $\angle{ADB}$ which is subtended by $\overline{AB}$ to get $\overline{BG}$.”
Since we’ve figured out pretty much everything in that figure besides$\overline{BG}$, it should be possible . And this is where we invoke Ptolemy’s theorem. Notice that ABGD is a quadrilateral. We have three of the four sides. One is the diameter, one is a given1, and the other was found from the Pythagorean theorem.
Similarly, to use Ptolemy’s theorem, we need to know the two diagonals. One was another “given” and the other was gotten from the Pythagorean theorem. Thus, the only unknown is that last side, $\overline{BG}$, which is what Ptolemy wants.
Before taking an example, let’s take the equation of Ptolemy’s theorem and solve it for the unknown:
$$\overline{AD} \cdot \overline{BG} + \overline{DG} \cdot \overline{AB} = \overline{BD} \cdot \overline{AG}$$
$$\overline{AD} \cdot \overline{BG} = \overline{BD} \cdot \overline{AG} – \overline{DG} \cdot \overline{AB}$$
$$\overline{BG} = \frac{\overline{BD} \cdot \overline{AG} – \overline{DG} \cdot \overline{AB}}{\overline{AD}}$$
Notice that the numerator has some subtraction in it. This corollary is often called the “chord of the difference” because we’re going to find a new chord that is the chord subtended by the difference in angles of two other known chords.
This will probably make more sense if I use an example and draw it on our diagram. We’ll take two chords of known lengths and angles from our table we started working on in the last post:
Making sure we know where everything came from:
We’ll make $\angle{ADB}$ be 60º which means its corresponding chord, $\overline{AB}$, is 60 parts in length. Similarly, we’ll make $\angle{ADG}$ be 72º and its chord, $\overline{AB}$ be 70;32,3 parts. Since $\overline{AD}$ is a diameter, we know its length must be 120 parts. We can find $\overline{BD}$ and $\overline{DG}$ from the Pythagorean theorem as stated above.
We also know $\angle{BDG}$ is 12º because it’s the difference in angles between 72º and 60º. Its subtended chord, $\overline{BG}$ is the unknown we’re looking for.
So we can start plugging into Ptolemy’s theorem we solved above:
$$\overline{BG} = \frac{(134;9,51 \cdot 70;32,3) – (139;11,40 \cdot 60)}{120}$$
$$\overline{BG} = 12;32,36$$
So we just got a new chord to add to our table: 12º = 12;32,36.
And it also means that we can start subtracting that from other angles to get even more chords as well as other combinations and it starts stacking up quickly. In particular, we could get the chord for 18º since we have 90º and 108º (or 72º and 90º) already and their difference is 18º. Then we have 18º and 12º which would give us 6º. Which we can repeatedly subtract from 180º (subtends an arc of 120 since its a diameter) and get every chord-angle pair from 0º to 180º.
So that little corollary ended up being pretty powerful.
But to get even finer divisions, we want to break up angles without necessarily knowing component parts. So Ptolemy develops another corollary to divide known angle-arc relationships in half.
In this new figure, there’s a few things to note:
First off, $\overline{AD}$ bisects $\angle{BAG}$ which makes sense since dividing known angles in half is what we’re shooting for.
Second, $\overline{DZ}$ is perpendicular to $\overline{AG}$.
Lastly, $\overline{AE} = \overline{AB}$
From these we can conclude that $\overline{BD} = \overline{DE}$. The reason for this is that $\triangle{ABD}$ and $\triangle{ADE}$ are identical, just flipped over the axis $\overline{AD}$. We know the triangles are identical because they share a common line ($\overline{AD}$), and we stated that $\overline{AE} = \overline{AB}$ and $\overline{AD}$ bisects $\angle{BAG}$ which means $\angle{BAD} = \angle{DAG}$. So we’ve got two sides equal as well as the enclosed angle which means the third side must be equal as well2.
However, both of these ($\overline{BD}$ and $\overline{DE}$) are also equal to $\overline{DG}$ because $\overline{AD}$ bisected the arc so the chords it produces ($\overline{BD}$ and $\overline{DG}$) must also be equal.
Since these two are equal legs of $\triangle{DEG}$, this fits the definition of an isoscles triangle. Since we started with $\overline{DZ}$ being perpendicular to its base, that means it’s dividing the base in half which means $\overline{EZ} = \overline{GZ}$.
Looking at $\overline{EG}$ we can see it’s the full length of $\overline{AG}$ with $\overline{AE}$ subtracted off. But as we stated above, $\overline{AE} = \overline{AB}$. So writing that another way,
$$\overline{EG} = \overline{AG} – \overline{AB}$$
And if we really want to isolate just a sub-piece of $\overline{EG}$ then we can state:
$$\overline{GZ} = \frac{1}{2} (\overline{AG} – \overline{AB})$$
We’ll come back to that in a second because Ptolemy now turns to another theorem from Euclid to state that $\triangle{ADG}$ is similar to $\triangle{DGZ}$ . And when you have similar triangles, that means their sides are proportional as well which allows us to state:
$$\frac{AG}{DG} = \frac{DG}{GZ}$$
Rearranging that algebraically and solving for one of the half chords we were after:
$$\overline{DG}^2 = \overline{AG} \cdot \overline{GZ}$$
$\overline{AG}$ is easy since it’s a diameter and $\overline{GZ}$ is that thing we just put on hold a minute ago. It’s time to substitute that in:
$$\overline{DG}^2 = \overline{AG} \cdot \frac{1}{2} (\overline{AG} – \overline{AB})$$
Taking stock once again, we now have what we’re after $\overline{DG}$ in terms of the diameter, and $\overline{AB}$. That’s nice, except $\overline{AB}$ isn’t the chord of the angle we’re breaking in half. However, we can get to the chord we’re after because $\overline{AB}$ is part of the right triangle $\triangle{ABG}$. The Pythagorean theorem tells us,
$$\overline{AB}^2 + \overline{BG}^2 = \overline{AG}^2$$
Solving that for the $\overline{AB}$ we want to get rid of:
$$\overline{AB} = \sqrt {\overline{AG}^2 – \overline{BG}^2}$$
Substituting that back into the equation we’re working on (and doing a wee bit of rearranging):
$$\overline{DG}^2 = \frac{1}{2} \overline{AG} \cdot (\overline{AG} – \sqrt {\overline{AG}^2 – \overline{BG}^2})$$
$$\overline{DG} = \sqrt{ \frac{1}{2} \overline{AG} \cdot (\overline{AG} – \sqrt {\overline{AG}^2 – \overline{BG}^2})}$$
It’s not especially pretty but it’s all in terms of only two things: diameter $\overline{AG}$ and the known chord we want to chop in half, $\overline{BG}$. And that’s Ptolemy’s second corollary, which finds the chord of half angles. And we didn’t actually even have to invoke Ptolemy’s theorem in this one.
So we can start diving previous angle-chord relationships in half, then divide the halves in half and so on to really start filling out our table. Whereas after the last corollary got us to 6º, we can cut 6º in half to get 3º, that in half to get $1 \frac{1}{2}$º, and that in half to get $\frac{3}{4}$º. We could go on, but Ptolemy is only shooting for every $\frac{1}{2}$º so we’ll quit there.
Recapping these corollaries, we’ve now been able to subtract angles and figure out the chord of the difference and divide an angle in half and get the resulting chord. But it sure would be nice to be able to add two angles and get the chord of that. After all, we’ve gotten to a point we have some pretty small angles so if we could just keep adding those together, we’d fill out a lot more of the table.
As usual, the setup:
$\overline{AD}$ and $\overline{BE}$ are diameters. Z is the center.
Again, we’ll be assuming some of the chords are known from previously derived chords, in particular, $\overline{AB}$ and $\overline{BG}$, which means the chord we’ll be trying to solve for is $\overline{AG}$.
Notice that we have two quadrilaterals in this figure: ABGD and BGDE. That means we can apply Ptolemy’s theorem. Starting with the latter:
$$\overline{BG} \cdot \overline{EG} = \overline{BG} \cdot \overline{DE} + \overline{BE} \cdot \overline{DG}$$
Let’s go through this term by term and see what we have or can find.
$\overline{BG}$ is one of our previously known chords.
$\overline{EG}$ is part of $\triangle{BGE}$ and since another side of that triangle is$\overline{BE}$ which is a diameter, it’s a right triangle and we can solve for $\overline{EG}$ via the Pythagorean theorem.
$$\overline{EG}^2 + \overline{BG}^2 = \overline{BE}^2$$
$$\overline{EG} = \sqrt{ \overline{BE}^2 – \overline{BG}^2}$$
Similarly, $\overline{DE}$ is part of $\triangle{BDE}$. Good news is that $\overline{BE}$ is a diameter, but we don’t know $\overline{BD}$ so we can’t solve for $\overline{DE}$ just yet. However, $\overline{BD}$ is part of $\triangle{ABD}$ where we again will have one of the sides as a previously derived chord, and another side that’s a diameter which means we can solve for $\overline{BD}$:
$$\overline{AB}^2 + \overline{BD}^2 = \overline{AD}^2$$
$$\overline{BD} = \sqrt{ \overline{AD}^2 – \overline{AB}^2}$$
With that in hand we can again apply the Pythogrean theorem to solve for $\overline{DE}$ in turn:
$$\overline{BD}^2 + \overline{DE}^2 = \overline{BE}^2$$
$$ \overline{AD}^2 – \overline{AB}^2 + \overline{DE}^2 = \overline{BE}^2$$
$$ \overline{DE}^2 = \overline{BE}^2 – \overline{AD}^2 + \overline{AB}^2$$
$$ \overline{DE} = \sqrt{\overline{BE}^2 – \overline{AD}^2 + \overline{AB}^2}$$
$\overline{BE}$ is a diameter.
Which means $\overline{DG}$ is the only one left and thus, can be solved for.
$$\overline{BG} \cdot \sqrt{ \overline{BE}^2 – \overline{BG}^2} = \overline{BG} \cdot \sqrt{\overline{BE}^2 – \overline{AD}^2 + \overline{AB}^2} + \overline{BE} \cdot \overline{DG}$$
$$\overline{BG} \cdot \sqrt{ \overline{BE}^2 – \overline{BG}^2} – \overline{BG} \cdot \sqrt{\overline{BE}^2 – \overline{AD}^2 + \overline{AB}^2} = \overline{BE} \cdot \overline{DG}$$
$$ \overline{DG} = \frac{\overline{BG} \cdot \sqrt{ \overline{BE}^2 – \overline{BG}^2} – \overline{BG} \cdot \sqrt{\overline{BE}^2 – \overline{AD}^2 + \overline{AB}^2}}{ \overline{BE}}$$
Yeah. That’s getting kinda messy. And there’s no good way to clean it up. And worse, it’s not what we were after. We were looking for $\overline{AG}$. So time to use Ptolemy’s theorem on the other quadrilateral.
$$\overline{AG} \cdot \overline{BD} = \overline{AB} \cdot \overline{DG} + \overline{BG} \cdot \overline{AD}$$
Taking stock again:
$\overline{AG}$ is the unknown we’re hoping we can solve for.
$\overline{BD}$ we stated above can come from being part of $\triangle{ABD}$ and we determined to be $\sqrt{ \overline{AD}^2 – \overline{AB}^2}$
$\overline{AB}$ is a previously known chord.
$\overline{DG}$ is what we solved for via our last application of Ptolemy’s theorem.
$\overline{BG}$ is a previously known chord.
$\overline{AD}$ is a diameter.
So again, we have all the pieces except $\overline{AG}$ so we can definitely solve for it. And I have. But it gets really messy and sufficiently wide that it actually breaks the page formatting. So for now it suffices to say that it can be done. Which is what Ptolemy did. In truth, he didn’t even show any of the math this time. Rather, he just walked through the logic quickly and said it can be done, in the same way I’m walking through all these theorems but not going to calculate every angle from 0-180º in $\frac{1}{2}$º intervals.
But we’ve shown so far that you can subtract chords to get new chords, divide chords, and now add chords. Unfortunately, none of this gets us quite down to the $\frac{1}{2}$º interval Ptolemy is shooting for. Nothing we’ve come up so far allows us to subtract or bisect our way down to it.
Which means we’re going to need one final push. And this one’s going to be a bit different because it’s not a cut and dry proof. Ptolemy says there’s no perfect way to get down to $\frac{1}{2}$º, so the best we can do is approximate it which we’ll get into next time via something known as Aristarchus’ inequality.
External References:
- We haven’t said what the value of these “given” chords are just yet, but remember that we’ve started putting together a table of chords that we could choose from.
- This is the Side-Angle-Side theorem at work.